marie wrote:as part of a ds problem I had to resolve:

x^2 > 2x i.e., {x squared is greater than 2x}

is the answer: x>2 and x<0 or x>2 and x<-2??

thxs,
please show all steps!

x^2 > 2x

x^2 - 2x > 0
x(x - 2) > 0

0 > x > 2

marie wrote:as part of a ds problem I had to resolve:

x^2 > 2x i.e., {x squared is greater than 2x}

is the answer: x>2 and x<0 or x>2 and x<-2??

thxs,
please show all steps!

As JeffB notes, we can simplify to:

x(x-2)>0

Let's see how we get from there to the final answer.

First, we recognize that to obtain a positive product, either both terms are positive or both are negative. Accordingly, we can set up two pairs of inequalities.

The positive case gives us:

x>0 AND x-2>0
x>0 AND x>2

When we have two inequalities pointing in the same direction, we must satisfy the more limiting inequality. In this case, that means that x>2.

The negative case gives us:

x<0 AND x-2<0
x<0 AND x<2

Again choosing the more limiting inequality, we get x<0.

So, to satisfy the original inequality, either:

x<0 OR x>2.

Stuart,
Thank you for that. Help me understand where the following line of reasoning goes wrong.

x^2 > 2x
assume x is negative therefore
(-x)(-x) > 2(-x)
divide both sides by -x
-x < 2
multiply by -1
x > -2

Thxs

marie wrote:Stuart,
Thank you for that. Help me understand where the following line of reasoning goes wrong.

x^2 > 2x
assume x is negative therefore
(-x)(-x) > 2(-x)
divide both sides by -x
-x < 2
multiply by -1
x > -2

Thxs

Assuming x is negative doesn't mean we replace it with "-x" in the equation; if x is negative and you write "-x", you're actually looking at a double negative, which is positive.

If we assume x is negative, we leave it as written, we just remember to flip the inequality when we divide by x. So:

x*x>2x

dividing by x:

x<2

which is trivial - all we've shown is that if x is negative, then it's less than 2, which is true of all negative numbers.

I guess the more fundamental issue is why you're assuming that x is negative - how does that help you solve this problem?